The standard enthalpy of formation of NH3(g) at 298 K is -46 kJ mol-1. Which of the following values is CORRECT for the standard enthalpy change for the following reaction? 2NH3(g) → N2(g) + 3H2(g) Select one: a. +92 kJ/mol b. -92 kJ/mol c. +46 kJ/mol d. -46 kJ/mol
The Correct Answer and Explanation is:
The correct answer is: a. +92 kJ/mol
Explanation:
The standard enthalpy of formation (ΔH°f) of a compound is the enthalpy change when 1 mole of a substance is formed from its elements in their standard states under standard conditions (298 K and 1 atm).
For ammonia (NH₃), the standard enthalpy of formation is given as:ΔHf∘[NH3(g)]=−46 kJ/mol\Delta H^\circ_f [\text{NH}_3(g)] = -46 \, \text{kJ/mol}ΔHf∘[NH3(g)]=−46kJ/mol
This means the reaction:12N2(g)+32H2(g)→NH3(g)\frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g)21N2(g)+23H2(g)→NH3(g)
has a ΔH° = -46 kJ/mol, indicating it is exothermic.
Now, consider the reaction in question:2NH3(g)→N2(g)+3H2(g)2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g)2NH3(g)→N2(g)+3H2(g)
This is the reverse of the formation reaction, but for 2 moles of ammonia instead of 1. Since formation of 1 mole of NH₃ releases 46 kJ, the formation of 2 moles would release:2×(−46)=−92 kJ2 \times (-46) = -92 \, \text{kJ}2×(−46)=−92kJ
Reversing the reaction (breaking NH₃ back into N₂ and H₂) will absorb the same amount of energy:ΔH∘=+92 kJ\Delta H^\circ = +92 \, \text{kJ}ΔH∘=+92kJ
This is because enthalpy is a state function—reversing a reaction changes the sign of ΔH but not its magnitude. Also, scaling the coefficients by 2 multiplies the enthalpy change by 2.
Final Answer:
+92 kJ/mol\boxed{+92 \, \text{kJ/mol}}+92kJ/mol
This is an endothermic reaction, meaning energy must be supplied to break down ammonia into nitrogen and hydrogen gases.
